In the figure, HJKL is a trapezium and HLMQ is a rhombus. HL is parallel to JK and LM = LN. QLK and LPN are straight lines. ∠LHQ = 42° and ∠PLQ = 28°. Find
- ∠HLK
- ∠LNM
(a)
∠HLQ
= (180° - 42°) ÷ 2
= 69° (Isosceles triangle)
∠HLK
= 180° - 69°
= 111° (Angles on a straight line)
(b)
∠NLM
= 69° - 28°
= 41°
∠LNM
= (180° - 41°) ÷ 2
= 69.5 ° (Isosceles triangle)
Answer(s): (a) 111°; (b) 69.5°